Question
Physics Question on laws of motion
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms-1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is: (g =10 ms-2)
A
23000
B
20000
C
34500
D
23500
Answer
34500
Explanation
Solution
Total force needed to go at a steady speed,
Force is equal to weight plus friction.
F = 20,000 + 3,000
F equals 23000. N
Power is equal to F ⋅ V.
= 23000 × 1.5
P is 34500. W
The correct answer is (C).34500.